Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $a = \dfrac{-p^2 - 6p}{3p^3 + 6p^2 - 144p} \div \dfrac{3p - 12}{5p^2 + 20p - 160} $
Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{-p^2 - 6p}{3p^3 + 6p^2 - 144p} \times \dfrac{5p^2 + 20p - 160}{3p - 12} $ First factor out any common factors. $a = \dfrac{-p(p + 6)}{3p(p^2 + 2p - 48)} \times \dfrac{5(p^2 + 4p - 32)}{3(p - 4)} $ Then factor the quadratic expressions. $a = \dfrac {-p(p + 6)} {3p(p + 8)(p - 6)} \times \dfrac {5(p + 8)(p - 4)} {3(p - 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-p(p + 6) \times 5(p + 8)(p - 4) } { 3p(p + 8)(p - 6) \times 3(p - 4)} $ $a = \dfrac {-5p(p + 8)(p - 4)(p + 6)} {9p(p + 8)(p - 6)(p - 4)} $ Notice that $(p + 8)$ and $(p - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-5p\cancel{(p + 8)}(p - 4)(p + 6)} {9p\cancel{(p + 8)}(p - 6)(p - 4)} $ We are dividing by $p + 8$ , so $p + 8 \neq 0$ Therefore, $p \neq -8$ $a = \dfrac {-5p\cancel{(p + 8)}\cancel{(p - 4)}(p + 6)} {9p\cancel{(p + 8)}(p - 6)\cancel{(p - 4)}} $ We are dividing by $p - 4$ , so $p - 4 \neq 0$ Therefore, $p \neq 4$ $a = \dfrac {-5p(p + 6)} {9p(p - 6)} $ $ a = \dfrac{-5(p + 6)}{9(p - 6)}; p \neq -8; p \neq 4 $